Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxMN = 101;
const int infty = 501;
int m, n, k, a[maxMN][maxMN], r[maxMN], c[maxMN];
void Enter() {
  cin >> m >> n;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j) cin >> a[i][j];
}
void Enter2() {
  m = 100;
  n = 100;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j) a[i][j] = 500;
  for (int i = 1; i <= m; ++i) a[i][99] = 20;
}
inline void Minimize(int& Target, int Value) {
  if (Target > Value) Target = Value;
}
void Init() {
  int cm;
  cm = infty;
  for (int i = 1; i <= m; ++i) Minimize(cm, a[i][1]);
  for (int i = 1; i <= m; ++i) r[i] = a[i][1] - cm;
  cm = infty;
  for (int j = 1; j <= n; ++j) Minimize(cm, a[1][j] - r[1]);
  for (int j = 1; j <= n; ++j) c[j] = a[1][j] - r[1] - cm;
}
int GetVal() {
  int res = a[1][1] - r[1] - c[1];
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i][j] - r[i] - c[j] != res) return -1;
  return res;
}
void Solve() {
  if (m < n)
    for (int i = 1; i <= m; ++i) r[i] += k;
  else
    for (int j = 1; j <= n; ++j) c[j] += k;
  int sum = 0;
  for (int i = 1; i <= m; ++i) sum += r[i];
  for (int j = 1; j <= n; ++j) sum += c[j];
  cout << sum << '\n';
  for (int i = 1; i <= m; ++i)
    for (; r[i] > 0; --r[i]) cout << "row " << i << '\n';
  for (int j = 1; j <= n; ++j)
    for (; c[j] > 0; --c[j]) cout << "col " << j << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  Enter();
  Init();
  k = GetVal();
  if (k < 0)
    cout << -1;
  else
    Solve();
}